What is area y= 1/(sqrt(4-9x^2)),(0, 1/3) ?

The answer to area y= 1/(sqrt(4-9x^2)),(0, 1/3) is pi/(18)

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area y= 1/(sqrt(4-9x^2)),(0, 1/3)

area

y = \frac{1}{4 - 9 x ^{2}}, (0, \frac{1}{3})

\frac{π}{18}

Decimal Notation

0.17453 \dots

Solution steps

Apply the area formula:

\int_{0}^{\frac{1}{3}} \frac{1}{4 - 9 x ^{2}} d x

Solve

\int_{0}^{\frac{1}{3}} \frac{1}{4 - 9 x ^{2}} d x : \frac{π}{18}

The area is:

= \frac{π}{18}

\frac{d}{d x} (- 7 sin (cos (x)))

\frac{\partial}{\partial x} (\frac{4 x - 4 z}{y + 4 z})

(2 x + 2 y)^{3} = 8 x^{3} + 8 y^{3}, (- 1, 1)

\int (2 x + 1) (x - 3) d x

\frac{\partial}{\partial x} (tan (x y^{3}))

What is area y= 1/(sqrt(4-9x^2)),(0, 1/3) ?
The answer to area y= 1/(sqrt(4-9x^2)),(0, 1/3) is pi/(18)